Difference between revisions of "Derivative of arccot"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> <strong>Theorem:</strong> $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(...") |
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| − | <strong>[[Derivative of arccot|Theorem]]:</strong> | + | <strong>[[Derivative of arccot|Theorem]]:</strong> The following formula holds: |
| − | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1}$$ | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ |
| + | where $\mathrm{arccot}$ denotes the [[arccot|inverse cotangent]] function. | ||
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<strong>Proof:</strong> If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | <strong>Proof:</strong> If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use [[implicit differentiation]] with respect to $z$ to get | ||
Revision as of 21:38, 15 May 2016
Theorem: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ where $\mathrm{arccot}$ denotes the inverse cotangent function.
Proof: If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use implicit differentiation with respect to $z$ to get $$-\csc^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ as was to be shown. █
